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2 years ago

What is the pH of 5.0 L of an aqueous solution that contains 1.0 grams of HBr and 1.0 grams of nitric acid

Chemistry

1 answer:

Nady [450]2 years ago

8 0

Hello!:

Molar mass:

HBr = 80.91 g/mol

HNO3 = 63.01 g/mol

Number of moles:

moles HBr = 1.0 / 80.91 => 0.01235 moles of HBr

moles HNO3 = 1.0 / 63.01 => 0.01587 moles of HNO3

So , total moles of H⁺ :

0.01235 + 0.01587 => 0.02822 moles of H⁺

Molarity of solution:

M = n / V

M = 0.02822 / 5.0

M = 0.005644

Therefore:

pH = - log [ H⁺ ]

pH = - log [ 0.005644 ]

pH = 2.25

Hope that helps!

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write the balanced equation for the reaction that occurs when butane undergoes combustionWhat is the coefficient of oxygen in th

love history [14]

Answer:

2C4H10 + 13O2 ----> 4CO2 + 10H2O

The coefficient of oxygen in the balanced equation is 13

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3 years ago

What is the highest energy to which doubly ionized helium atoms (alpha particles) can be accelerated in a DC accelerator with 3

pentagon [3]

Answer:

Highest energy will be equal to $9.6\times 10^{-13}J$

Explanation:

Charged on doubly ionized helium atom $q=2e=2\times 1.6\times 10^{-19}C=3.2\times 10^{-19}C$

It is accelerated with maximum voltage of 3 MV

So voltage $V=3\times 10^6volt$

Now energy is given by $E=qV=3.2\times 10^{-19}\times 3\times 10^6=9.6\times 10^{-13}J$

So highest energy will be equal to $9.6\times 10^{-13}J$

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2 years ago

Consider the following reaction where K. = 154 at 298 K: 2NO(g) + Brz(9) 2NOBr(g) A reaction mixture was found to contain 2.69x1

bekas [8.4K]

Explanation:

$2NO(g) + Br_2(g)\rightleftharpoons 2NOBr(g)$

Equilibrium constant of reaction = $K=154$

Concentration of NO = $[NO]=\frac{2.69\times 10^{-2} mol}{1 L}=2.69\times 10^{-2} M$

Concentration of bromine gas = $[Br_2]=\frac{3.85\times 10^{-2} mol}{1 L}=3.85\times 10^{-2} M$

Concentration of NOBr gas = $[Br_2]=\frac{9.56\times 10^{-2} mol}{1 L}=9.56\times 10^{-2} M$

The reaction quotient is given as:

$Q=\frac{[NOBr]^2}{[NO]^2[Br_2]}=\frac{(9.56\times 10^{-2} M)^2}{(2.69\times 10^{-2} M)^2\times 3.85\times 10^{-2} M}$

$Q=328.06$

$Q>K$

The reaction will go in backward direction in order to achieve an equilibrium state.

1. In order to reach equilibrium NOBr (g) must be produced. False

2. In order to reach equilibrium K must decrease. False

3. In order to reach equilibrium NO must be produced. True

4. Q. is less than K . False

5. The reaction is at equilibrium. No further reaction will occur. False

8 0

2 years ago

If 42.8 mL of 0.204 M HCl solution is needed to neutralize a solution of Ca(OH)2, how many grams of Ca(OH)2 must be in the solut

Aneli [31]

Hey There!

At neutralisation moles of H⁺ from HCl = moles of OH⁻ from Ca(OH)2 so :

0.204 * 42.8 / 1000 => 0.0087312 moles

Moles of Ca(OH)2 :

2 HCl + Ca(OH)2 = CaCl2 + 2 H2O

0.0087312 / 2 => 0.0043656 moles ( since each Ca(OH)2 ives 2 OH⁻ ions )

Therefore:

Molar mass Ca(OH)2 = 74.1 g/mol

mass = moles of Ca(OH)2 * molar mass

mass = 0.0043656 * 74.1

mass = 0.32 g of Ca(OH)2

Hope that helps!

6 0

3 years ago

Determine the mass in grams of each nacl solution that contains 1.5g of nacl

Rama09 [41]

Answer:- 3333 g of solution.

Some of the question part is missing here. It would be like, "Determine the mass in grams of each NaCl solution that contains 1.5 g of NaCl.

(i) 0.045% NaCl by mass

Solution:- 0.045% NaCl by mass means 0.045 g of NaCl are present in 100 g of solution. 1.5 g of NaCl would be present in how many grams of solution?

We could solve this using proportions...

(0.045/100) = (1.5/X)

0.045(X) = 1.5(100)

0.045X = 150

X = 150/0.045 = 3333

So, 1.5 g of NaCl is present in 3333 g of solution.

7 0

3 years ago